Indefinite integral: ∫ [f (x) / F ′ (x) - F 2 (x) f ″ (x) / F ′ (x) 3] DX

Indefinite integral: ∫ [f (x) / F ′ (x) - F 2 (x) f ″ (x) / F ′ (x) 3] DX

∫[ƒ(x)/ƒ'(x) - ƒ(x)²ƒ''(x)/ƒ'(x)³] dx=∫[f(x)/f′(x)]•(f′(x)²-ƒ(x)²ƒ''(x))/ƒ'(x)²)dx=∫[f(x)/f′(x)]•(f(x)/ƒ'(x))′d...

Integral: DX

Sin (x ^ 2) is not equal to (SiNx) ^ 2! Some people are very hard
D rivers and mountains have water

Find the indefinite integral ∫ sin (2x) / (1 + cosx) DX

(1 + cosx) = - 2 ∫ cosxd (cosx) / (1 + cosx x) = - 2 ∫ cosxd (cosx) / (1 + cosx x) = - 2 ∫ cosxd [ln (1 + cosx x)] using the integral method, get the next step = - 2cosxln (1 + cosx) + 2 ∫ ln (1 + cosx) dcosx = - 2cosxln (1 + cosx) + 2 ∫ ln (1 + cosx) dcosx = - 2cosxln (1 + cosx) + 2 ? 2 ? 8747; 8747; ln (1 + cosx) dcosx = - 2cosxln (1 + cosx ln (1 + cosx) d (1 + cosx) this step

∫ (sin √ X / √ x) DX for indefinite integral

∫（sin√x/√x）dx
= -2 ∫（sin√x）d√x
= 2 cos√x + c

Find the indefinite integral ∫ cosx / (sin ^ 3) x

∫cosx/(sin^3)x dx=∫dsinx/(sin^3)x=(sinx)^(-2)/(-2)+C=-1/(2sinx^2)+C

Sin (x ^ 1 / 2) DX for indefinite integral

cos（x^1/2）*（x^(-1/2))/2+C
C is added to all indefinite integrals

Given the function f (x) = (a ^ x) - (4-A ^ x) - 1 (a > 0, and a is not equal to 1), find the definition domain and value range of function f (x) Do it with the substitution method!

When 0 < a < 1, 4 - A ^ x > 0, the solution is: x > loga ^ 4; when a > 1, 4 - A ^ x > 0, the solution is: x < loga ^ 4; then: 0 ＜√ (4 - A ^ x) < 2, let √ (4 - A ^ x) = m, then 0 ≤ m ≤ 2. F (m) = - m ^ 2 - 2m + 3, m ∈ (0,2)

It is proved that the function f (x) = - [(root a) / (a ^ x + root a)] (a is greater than 0 and not equal to 1) It is proved that the graph of the function f (x) = - [(root a) / (a ^ x + root a)] (a is greater than 0 and not equal to 1) is symmetric about the point (1 / 2, - 1 / 2)

It is very simple to prove the centrosymmetry. Let X1 (x, y) on f (x) and the symmetric point of X1 about the symmetry center be (x0, Y0). As long as (x0, Y0) is on f (x), then f (x) is symmetric about the point (1 / 2, - 1 / 2). The formula of midpoint coordinate is: x0 = 2.0.5-x, Y0 = 2 · (- 0.5) - Y; as long as it is proved that f (x0) = Y0

The known function f (x) = root x divided by (A's x power + root a). (a > 0 and a is not equal to 1) (1) Proof: the image of the function y = f (x) is symmetric about the point (half, half) (2) Find the value of F (- 2) + F (- 1) + F (0) + F (1) + F (2) + F (3)

The original question is wrong
F (x) = root a divided by (x power of a + root a)
=√a/(a^x+√a)
To prove that the image of the function y = f (x) is symmetric about a point (1 / 2,1 / 2),
It is only necessary to prove that f (x) + F (1-x) = 1
It is proved as follows:
Because f (x) = √ A / (a ^ x + √ a)
So f (x) + F (1-x) = √ A / (a ^ x + √ a) + √ A / (a ^ (1-x) + √ a)
The numerator and denominator of the second term is multiplied by a ^ X
=√a/(a^x+√a) + √a* a^x/(a+√a* a^x)
=√a*√a/(√a* a^x + a) + √a* a^x/(a+√a* a^x)
=a/(√a* a^x + a) + √a* a^x/(a+√a* a^x)
=( a+ √a* a^x) /(a+√a* a^x)=1,
So the conclusion holds
Because f (x) + F (1-x) = 1
So f (- 2) + F (- 1) + F (0) + F (1) + F (2) + F (3)
=[ f(-2) +f(3)]+[ f(-1)+f(2)]+[ f(0)+f(1)]
=1+1+1=3.

Known function f (x)= 3−ax A − 1 (a ≠ 1). If f (x) is a decreasing function on the interval (0, 1], then the value range of real number a is______ .

f′（x）=−a
2(a−1)
3−ax；
If f (x) is a decreasing function on the interval (0, 1], then f '(x) < 0;
That is − a
If a − 1 < 0, a < 0 or a > 1 can be obtained;
And 3-ax ≥ 0, that is, a ≤ 3
x. On (0, 1], hang holds, 3
The minimum value of X on (0, 1] is 3,  a ≤ 3;
The value range of real number a is (- ∞, 0) ∪ (1,3]
So the answer is: (- ∞, 0) ∪ (1, 3]